3 MATH 45, PRACTICE FINAL EXAM SOLUTIONS. 3 a We need to prove uniqueness of solutions for the boundary value problem. It suffices to show that, if w solves: w = on Ω w = on Ω. then w = on Ω. This claim follows immediately from the Maximum principle for solutions to the Laplace equation. In Homework 6, Exercise, we saw how to deduce this claim by using the Energy method. b We recall that, on R 3, it is true that: Hence: This quantity is non-negative if x = x + x + x 3 = 6. u + λx = u + λ x = f + 6λ. λ max Ω f. 6 For such a λ, the function u + λx is subharmonic. c Let us take λ as in part b. We note that the function u + λx solves the following boundary value problem: u + λx = f + λx on Ω u + λx = g + λx on Ω. By using the Maximum principle for subharmonic functions Exercise 3 from Homework 6, it follows that for all x Ω, the following bound holds: ux + λx max gy + λy = gy + λy y Ω for some y Ω. In particular, since λ : ux gy + λy x gy + λy. Since Ω is bounded, we can find a constant K > depending only on Ω such that for all y Ω, one has: y K. Hence, it follows that for all x Ω: ux max Ω g + λ K. We now note that, if f, we can take λ = and the claim will follow from the maximum principle for subharmonic functions. If it is not the case that f, we can take λ = max Ω f 6 >, and C = K 6 to obtain that for all x Ω: ux max g + C max f. Ω Ω We note that the constant C > depends only on the domain Ω. Exercise 4. The sign of the Green s function Suppose that Ω R 3 is a bounded domain. Given x Ω, we consider the Green s function Gx, x. We recall from the definitions that then the function: Hx, x := Gx, x + 4π x x

4 4 MATH 45, PRACTICE FINAL EXAM SOLUTIONS. is well-defined and harmonic on all of Ω. a Show that, for all x Ω \ x }: Gx, x <. b Does the claim still hold if we assume that Ω is a domain in two-dimensions? a We can assume without loss of generality that Ω is connected. If this is not the case, we just consider each connected component of Ω separately. Let us fix ɛ > small. We define: Ω ɛ := Ω \ Bx, ɛ. If ɛ is sufficiently small, then Ω ɛ is also connected. Let us note that the function Gx, x = Hx, x 4π x x is harmonic on Ω ɛ. We recall that Hx, x is bounded for x Ω and that lim x x x x = +. In particular, it follows that if we choose ɛ > sufficiently small, we obtain that: Gx, x < for x Bx, ɛ. By construction of the Green s function, we know that: Gx, x = for x Ω. We no use the weak Maximum principle for the harmonic function Gx, x on Ω ɛ in order to deduce that: Gx, x for x Ω ɛ. We can say even more if we use the Strong Maximum principle. Namely, since Gx, x is not identically equal to zero on Ω ɛ, it follows that: We can now let ɛ in order to deduce that: Gx, x < for x Ω ɛ. Gx, x < for x Ω \ x }. b In two dimensions, we recall from Exercise on Homework 8 that, in the definition of the Green s function, the function Hx, x := Gx, x π log x x has to be harmonic. The same argument as above applies since we know that lim x x log x x =. In particular, we obtain that Gx, x < for x Ω \ x }. Exercise 5. Finite speed of propagation for the wave equation without using the d Alembert formula Suppose that u : R x R t solves the initial value problem for the wave equation: u tt c u xx = on R x R t u t= = φ, u t t= = ψ on R x. Given x R and t >, we define the cone: Kx, t := x, t R [, t ], x x ct t}. Furthermore, for t t, we define the local energy: et := x+ct t [ u t x, t + c u x x, t] dx. a Show that: x ct t

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